Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (c)/Solution 1

We have $f^{″}(x)=\frac{9}{7}\cdot \frac{1\cdot x^{\frac{5}{7}}-(x+1)\cdot \frac{5}{7x^{\frac{2}{7}}}}{x^{\frac{10}{7}}}=\frac{9}{7}\cdot \frac{7x-5(x+1)}{7x^{\frac{12}{7}}}=\frac{9(2x-5)}{49x^{\frac{12}{7}}}.$ This denominator will always be nonnegative (because of the even number $12$ in the exponent), so it suffices to understand when the numerator is positive/negative. Note also that $f^{″}(x)$ is undefined at $x=0$. The numerator is zero when $x=5/2$, positive when $x>5/2$, and negative when $x<5/2$. We therefore conclude that $f(x)\text{ is concave up on }(\frac{5}{2},+\mathrm{\infty })\text{ and concave down on }(-\mathrm{\infty },0)\cup (0,\frac{5}{2})$.